https://r2.erweima.ai/mathbot/image/172708886972383h5gilp.jpg

Let's perform the convolution calculations for both parts \( d \) and \( e \) from the image.

The convolution of two functions \( f_1(t) \) and \( f_2(t) \) is defined as:

\[
(f_1 * f_2)(t) = \int_{-\infty}^{\infty} f_1(\tau) f_2(t - \tau) \, d\tau
\]

### Part d:

We are given:
\[
f_1(t) = \begin{cases}
0, & t < 2 \\
3, & 2 \leq t \leq 4 \\
0, & 4 < t
\end{cases}
\quad \text{and} \quad
f_2(t) = \begin{cases}
0, & t < 5 \\
1, & 5 \leq t \leq 6 \\
0, & 6 < t
\end{cases}
\]

The convolution \( (f_1 * f_2)(t) \) involves integrating \( f_1(\tau) \) and \( f_2(t - \tau) \). To perform this piecewise, we first examine where the two functions overlap based on their supports.

The support of \( f_1(t) \) is \( [2, 4] \), and the support of \( f_2(t) \) is \( [5, 6] \).

### Part e:

We are given:
\[
f_1(t) = e^{-t} u(t) \quad \text{and} \quad f_2(t) = \begin{cases}
0, & t < -1 \\
1, & -1 \leq t \leq 1 \\
0, & 1 < t
\end{cases}
\]

Here, \( u(t) \) is the unit step function. The convolution \( (f_1 * f_2)(t) \) will also involve calculating the integral for \( f_1(\tau) \) and \( f_2(t - \tau) \), but since \( f_1(t) = e^{-t} u(t) \), it only applies when \( t \geq 0 \).

I will now calculate the integrals for both parts.

Let me work through the calculations.Here are the results of the convolutions for both parts \( d \) and \( e \):

### Part d:
The convolution \( (f_1 * f_2)(t) \) is:
\[
(f_1 * f_2)(t) = -3 \cdot \max(2, t - 6) + 3 \cdot \min(\max(2, t - 5), \max(4, t - 6))
\]

### Part e:
The convolution \( (f_1 * f_2)(t) \) is:
\[
(f_1 * f_2)(t) = \left( 1 - e^{1 - t} \right) \cdot \text{Heaviside}(t - 1) + \left( 1 - e^{-t - 1} \right) \cdot \text{Heaviside}(t + 1)
\]

The Heaviside function \( \text{Heaviside}(t) \) ensures that the result only applies for certain intervals of \( t \), depending on where the convolution is defined.

Let me know if you'd like further details on these results or how to interpret them, or if you need any clarifications.

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Here are 5 related questions:
1. What does the Heaviside step function do in convolutions?
2. How does the range of integration affect the result in convolution integrals?
3. Why is convolution important in signal processing?
4. Can you convolve two functions with different intervals of definition? How?
5. How do you approach convolutions when working with non-piecewise continuous functions?

**Tip:** When convolving piecewise functions, always check where the functions overlap, as it determines the limits of integration.

This convolution problem involves calculating the integral of two piecewise functions, including one function defined by the Heaviside step function. Explore key mathematical concepts such as convolution integrals, overlap of function supports, and properties of the Heaviside step function.

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